1) Vì a⊥d , b⊥d  ⇒ a // b

⇒\(\widehat{A_1}=\widehat{B}=80^o\) (ở vị trí so le trong)

⇒\(\widehat{A_3}=\widehat{B}=80^o\)(ở vị trí đồng vị)

Do  \(\widehat{A_2}+\widehat{B}=180^o\)

     (hai góc trong cùng phía)

Thay số:\(\widehat{A_2}+80^o=180^o\)

           ⇒\(\widehat{A_2}=100^o\)

2)a.Vì Ax//By⇒\(\widehat{A}=\widehat{ABy}=30^o\)

Mà \(\widehat{ABC}=70^o\)

\(\Rightarrow\widehat{yBC}=\widehat{ABC}-\widehat{CBy}=70^o-30^o=40^o\)

b. Xét Bx và Ct có :\(\widehat{CBy}=\widehat{C}=40^o\) là hai góc so le trong bằng nhau

⇒Bx//Ct  . Mà Ax//By

⇒Ax//Ct

a) Ta có: \(\widehat{BAC}+\widehat{xAC}=180^0\)(hai góc kề bù)

\(\Leftrightarrow\widehat{xAC}=100^0\)

\(\Leftrightarrow\widehat{xAt}=\widehat{CAt}=\dfrac{\widehat{xAC}}{2}=\dfrac{100^0}{2}=50^0\)

b) Ta có: \(\widehat{CAt}=\widehat{BCA}\left(=50^0\right)\)

mà hai góc này là hai góc ở vị trí so le trong

nên At//BC(Dấu hiệu nhận biết hai đường thẳng song song)

Đây là bài lớp 7 mà nhưng mik mới hc lớp 6 sr nha

Vì BC và Cx là 2 tia đối nên \(\widehat{BCA}\) và \(\widehat{ACx}\) là 2 góc kề bù

\(\Rightarrow\widehat{ACB}+\widehat{ACx}=180^o\)

     \(40^o+\widehat{ACx}=180^o\)

     \(\widehat{ACx}=140^o\)

b) Ta có:\(\widehat{ACB}+\widehat{ABC}+\widehat{BAC}=180^o\) (tổng 3 góc trong 1 tam giác)

             ​​​\(40^o+\widehat{ABC}+70^o=180^o\)​

            \(\widehat{ABC}=70^o\)(1)

Vì Oy là phân giác của \(\widehat{ACx}\) nên \(\widehat{xCy}=\dfrac{\widehat{ACx}}{2}=\dfrac{140^o}{2}=70^o\)(2)

Từ (1),(2) => \(\widehat{ABC}=\widehat{xCy}\)

c)Cặp góc đồng vị là \(\widehat{ABC}\) và \(\widehat{xCy}\)

\(5-\left|x-\dfrac{1}{2}\right|=3\)

\(\left|x-\dfrac{1}{2}\right|=5-3\)

\(\left|x-\dfrac{1}{2}\right|=2\)

=>\(x-\dfrac{1}{2}=2\)                      hoặc                      \(x-\dfrac{1}{2}=-2\)

  \(x=2+\dfrac{1}{2}\)                                                       \(x=-2+\dfrac{1}{2}\)

  \(x=\dfrac{5}{2}\)                                                            \(x=\dfrac{-3}{2}\)

Vậy...\(\left(x+\dfrac{1}{3}\right)^2+3=2\)

          \(\left(x+\dfrac{1}{3}\right)^2=2-3\)

          \(\left(x+\dfrac{1}{3}\right)^2=-1\)

Vi \(\left(x+\dfrac{1}{3}\right)^2\) luôn luôn lớn hơn 0 nên

\(\left(x+\dfrac{1}{3}\right)^2=-1\)(vô lý)

a) - Dùng quỳ tím:

+ Hóa đỏ -> dd H2SO4

+ Hóa xanh -> dd NaOH

+ Không đổi màu -> dd NaCl

b) - Dùng quỳ tím:

+ Hóa xanh -> dd Ca(OH)2

+ Hóa đỏ -> dd H2SO4, dd HCl

- Dùng dd BaCl2:

+ Có kết tủa trắng BaSO4 -> dd H2SO4 

+ Không có kt trắng -> dd HCl

PTHH: H2SO4 + BaCl2 -> BaSO4 (kt trắng) + 2 HCl

c) - Dùng quỳ tím:

+ Hóa xanh : dd KOH 

+ Hóa đỏ : dd HNO3, dd HCl

- Dùng dd AgNO3:

+ Có kt trắng AgCl -> dd HCl

PTHH: HCl + AgNO3 -> AgCl (trắng) + HNO3

+ Còn lại ko ht -> dd AgNO3

1.

Với \(n=0;1\) không thỏa mãn

Với \(n>1\)

\(A=\left(n^2+n\right)^2+n^2+3n+7>\left(n^2+n\right)^2\)

\(A=\left(n^2+n+2\right)^2-\left[3\left(n^2-1\right)+n\right]< \left(n^2+n+2\right)^2\)

\(\Rightarrow\left(n^2+n\right)^2< A< \left(n^2+n+2\right)^2\)

\(\Rightarrow A=\left(n^2+n+1\right)^2\)

\(\Rightarrow n^4+2n^3+2n^2+3n+7=\left(n^2+n+1\right)^2\)

\(\Rightarrow n^2-n-6=0\Rightarrow\left[{}\begin{matrix}n=-2\left(loại\right)\\n=3\end{matrix}\right.\)

3.

TH1:

\(x>y\Rightarrow x^2y^2+x-y>x^2y^2\)

Mặt khác x; y nguyên dương \(\Rightarrow y\ge1\Rightarrow xy-\left(x-y\right)=x\left(y-1\right)+y>0\Rightarrow xy>x-y\)

\(\Rightarrow2xy+1>x-y\Rightarrow x^2y^2+x-y< x^2y^2+2xy+1\)

\(\Rightarrow x^2y^2< x^2y^2+x-y< \left(xy+1\right)^2\)

\(\Rightarrow x^2y^2+x-y\) nằm giữa 2 SCP liên tiếp nên ko thể là SCP (trái giả thiết) \(\Rightarrow\) loại

TH2: \(x< y\Rightarrow x^2y^2+x-y< x^2y^2\)

\(x-y-\left(-2xy+1\right)=\left(x-1\right)+y\left(2x-1\right)>0\Rightarrow x-y>-2xy+1\)

\(\Rightarrow x^2y^2+x-y>x^2y^2-2xy+1=\left(xy-1\right)^2\)

\(\Rightarrow\left(xy-1\right)^2< x^2y^2+x-y< x^2y^2\)

\(\Rightarrow x^2y^2+x-y\) nằm giữa 2 SCP liên tiếp \(\Rightarrow\) ko thể là SCP => trái giả thiết => loại

Vậy \(x=y\)

2.

\(C=n^4-2n^3+3n^2-6n+8\)

\(C=\left(n^2-n\right)^2+2n^2-6n+8=\left(n^2-n\right)^2+2\left(n-\dfrac{3}{2}\right)^2+\dfrac{7}{2}>\left(n^2-n\right)^2\)

\(C=\left(n^2-n+3\right)^2-\left(4n^2+1\right)< \left(n^2-n+3\right)^2\)

\(\Rightarrow\left(n^2-n\right)^2< C< \left(n^2-n+3\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}C=\left(n^2-n+1\right)^2\\C=\left(n^2-n+2\right)^2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}4n-7=0\left(vn\right)\\2n^2+2n-4=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}n=1\\n=-2\end{matrix}\right.\)

refer

1/ I like table tennis and that’s why I spend a lot of time on that game.
=>I spend a lot of time on table tennis because I like it.
2/ Because the weather was cold, we had to cancel our picnic.
=> The weather was cold, so we had to cancel our picnic.
3/ I won’t forget to phone her. You know, I like her so much.
=> You know I like her so much so I won't forget to phone her
4/ They tried their best to complete the course and that was why they passed.
=> they tried their best to complete the course, so they passed
5/ I love my city so much, so I think I will stay and work here in my city.
=> I think I will stay and word in my city because I love it so much.

1 Because I like table tennis, I spend a lot of time on that game

2 The weather was so cold that we had to cancel our picnic

3 I won't forget to phone her because I like her so mich

4 They pass the course because they tried their best to complete the course

5 Because I love my city so much, I think I will stay and work here in my city

1 Because I like table tennis, I spend a lot of time on that game

2 The weather was so cold that we had to cancel our picnic

3 I won't forget to phone her because I like her so mich

4 They pass the course because they tried their best to complete the course

5 Because I love my city so much, I think I will stay and work here in my city

3.B (thì HTHT với dấu hiệu yet)

4. B

5. B (wish S Ved: ước trong iện tại)

6. D (thì QKĐ với dấu iệu Last night)